Answered

Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

a 100 kg gymnast comes to a stop after tumbling. her feet do -5000J of net work to stop her. Use the work-kinetic energy theorem to find the girl's initial velocity when she began to stop?

Sagot :

W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j 
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s

Answer:

10 m/s

Explanation:

The work-kinetic energy theorem states that the net work done on an object is equal to the change in the kinetic energy of the object.

In formula:

[tex]W=K_f -K_i[/tex] (1)

where

W is the work done

Ki is the initial kinetic energy

Kf is the final kinetic energy

In this problem, we have:

[tex]W=-5000 J[/tex] the net work done on the gymnast

[tex]m=100 kg[/tex] is the mass of the gymnast

[tex]v_f = 0[/tex] is the final velocity of the gymnast, so her final kinetic energy is also zero:

[tex]K_f = \frac{1}{2}mv_f^2 = 0[/tex]

Therefore, we can rewrite eq.(1) as

[tex]W=-\frac{1}{2}mv_i^2[/tex]

where [tex]v_i[/tex] is the initial velocity of the girl. By substituting the numbers and re-arranging the equation, we find:

[tex]v_i = \sqrt{-\frac{2W}{m}}=\sqrt{-\frac{2(-5000 J)}{100 kg}}=10 m/s[/tex]

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.